The longitudinal waves starting from a ship return from the bottom of the sea to the ship after 2.64 s. If the bulk modulus of water be `220 mm^(-2)` and the density `1.1xx10^(3) kg m^(-3)`.calculate the depth of the sea. Take g =`9.8 N kg^(-1)`.

Answer

To solve the problem, we need to calculate the depth of the sea using the given information about the longitudinal waves. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Problem
The time taken for the wave to travel to the bottom of the sea and back to the ship is given as 2.64 seconds. Since the wave travels to the bottom and back, the time for the wave to reach the bottom is half of the total time.

### Step 2: Calculate the Time to Reach the Bottom
\[ t_{\text{bottom}} = \frac{2.64 \, \text{s}}{2} = 1.32 \, \text{s} \]

### Step 3: Calculate the Speed of Sound in Water
The speed of sound in a medium can be calculated using the formula:
\[ v = \sqrt{\frac{K}{\rho}} \]
where \( K \) is the bulk modulus and \( \rho \) is the density.

### Step 4: Convert Units
The bulk modulus is given as \( 220 \, \text{mm}^{-2} \). We need to convert this to \( \text{N/m}^2 \) (or Pascals):
\[ K = 220 \, \text{mm}^{-2} = 220 \times 10^6 \, \text{N/m}^2 \]
The density is given as \( 1.1 \times 10^3 \, \text{kg/m}^3 \).

### Step 5: Substitute Values into the Speed Formula
Substituting the values into the speed formula:
\[ v = \sqrt{\frac{220 \times 10^6 \, \text{N/m}^2}{1.1 \times 10^3 \, \text{kg/m}^3}} \]

### Step 6: Simplify the Calculation
Calculating the fraction:
\[ \frac{220 \times 10^6}{1.1 \times 10^3} = \frac{220}{1.1} \times 10^3 = 200 \times 10^3 = 2 \times 10^5 \]

Now, taking the square root:
\[ v = \sqrt{2 \times 10^5} = \sqrt{2} \times 10^{2.5} \approx 447.21 \, \text{m/s} \]

### Step 7: Calculate the Depth of the Sea
Using the speed of sound and the time to reach the bottom, we can find the depth:
\[ d = v \times t_{\text{bottom}} = 447.21 \, \text{m/s} \times 1.32 \, \text{s} \approx 591.14 \, \text{m} \]

### Step 8: Final Result
The depth of the sea is approximately:
\[ d \approx 591.14 \, \text{m} \]